Lie groups anyone?

Having bored you all rigid with group theory, let me now take you on a moreinteresting journey, to the Lie groups. (pronounce it as "Lee")

Definition A Lie group is a manifold endowed with groupproperties (of the sort we (!) have been discussing).So what is a "manifold". It is a topological space withcertain additional properties. So, you don't know what a topologicalspace is, huh? Quite, so let's get dirty. Consider a set S.

DefinitionA class T of subsets of S is said to a be topology on S iff thefollowing are true:finite intersection of elements of T (aka subsets of S) are in T;infinite union or elements of T (aka subsets of S) are in T;S is in T;Ø is in T.

The pair (S,T) is referred to as a topological space. (Note that it isusual to abuse notation hugely and assert, for example, that "X is atopological space" (when it is, of course), so take note: it means X =(S,T))

I'll give some examples in a minute, but let's press on a bit.The elements of T (aka subsets of S) are said to be the open sets in(S,T). Clearly, elements of (S,T) in the complement of T are the closedsets in (S,T).

So let's have a couple of examples. Consider the subsets S and Ø of S.The topology T = {Ø, S} is referred to as the indiscrete (or trivial,or concrete) topology on S.Now consider the power set on S, that is all elements and theircombinations. (The power set P(S) for any set with n elements, is 2ⁿe.g. for S = {a, b, c}, P(S) = {{a}, {b}, {c}, {a,b}, {a,c}, {b,c},{a,b,c}, Ø}). This is in fact a topology on S and is referred to as thediscrete topology on S: it is the finest possile topology on S.

Now remember this: elements in the topology T on S are subsets of S.But elements of S are points, say s and r. So T is a set of sets, so tospeak.

Let (S,T) be a topological space. For some point s in S, an open set in(S,T) containing s is referred to as a neighbourhood of s.Confused? Good. Here's something rather reassuring. The real line R (Itrust you all know what that is), where we do our merry LaTexing, is atopological space, is a manifold, is a ring, is a vector space is agroup........

There are a couple more things I need to say about topological spaces.

Let X and Y be topological spaces (note the abuse of notation Ireferred to earlier). A map f: X → Y is said to be continuous iff, foreach x in the domain of f and each y in the codomain, there is aneighbourhood U of Y (aka open set containing y) such that thepre-image f-¹(U) (containing x) is open in X. It is (relatively) easyto show that this corresponds to the usual ε - δ definition ofcontinuity we learn in school.

There are loads of properties of topological spaces Icould mention, but I think I only need these two before getting to manifolds.

DefinitionA topological space is said to be connected iff it is not theunion of non-empty disjoint sets. Effecitively this means we can roamover the space without ever falling into a hole.

DefinitionA topological space X is said to be Haudorff iff, for each pair ofpoints x, y in X there are open sets U containing x and V containing ysuch that U intersect V = Ø. Basically this is is saying that in thetopological space R, for example, any line connecting x and y can beinfinitely subdivided.

As always, I'll try to answer questions (though I am still learning this stuff myself)

Re: Lie groups anyone?

I've never seen this use of the word class. Is there a mathimatical definition behind it, or do you just mean "type"?

I don't think I made it up, but I use it to mean "a collection".

There are two words, disjoint and open, which you used. I know their meaning and everything, but I only know them from their use on the reals. Is there a more topological definition?

As far as disjoint, it means the same here as anywhere: two sets aredisjoint if they have no elements in common.

Regarding "open" this is slightly more subtle: a subset U of S is openiff it is included in a topology T on S. Note that subsets can be bothopen and closed or neither under this definition. To reassure you,however, in what is called the usual (or standard) topology on the real line (this is the union of open intervals in R), open sets are openintervals of the form (a, b), just as we always thought. Beware though, in other topologies on R, some open intervals may not be in the topology i.e. they are not open sets in R viewed as a topologicalspace.

Is it that any nonempty set S can be put into a topology X byjust picking the right set of subsets T?

Hm. I know what you mean, but strictly speaking elements of Tare subsets of S. But yes, often we can choose a topology in theway you suggest, but not always.

A topological space is said to be connected iff it is notthe union of non-empty disjoint sets.

Is this definition meaningless for finite topologies?

Do you ask this because I said that infinite union of open sets isopen? I guess I meant infinite number of unions, not the unionof an infinite number of sets So the answer is no, it's notmeaningless.

Can you give an example of a topological space that is notHaudorff?

Ha! Good question. Non-Hausdorff spaces tend to be trivial orpathological, although I can think of an important exception. Thetopology T on S denoted by T = {Ø, S} is not Hausdorff: all points in S are included in the same open set, namely S.

The important exception, which I didn't really want to get into (notyet anyway) is the quotient topology: for some equivalence relation ~on X, the topology that arises from a partition into equivalenceclasses of some Hausdorff space X need not be Hausdorff.

Re: Lie groups anyone?

OK,I want to say a bit more about maps between topological spaces, because they are important.

But first let me say a bit more about the Hausdorff property, as ittripped me up big time in some exercise I was given early on.

Let X be a topological space, and let x, y, z be points in X. Now X isHausdorff if I can find some subset A containing x and some subset Bcontaining y such that A ∩ B = Ø. Now let z be in the neighbourhood(aka open set) C.

It need not follow that A ∩ C = Ø, nor that B ∩ C = Ø. In other words,neighbourhoods of x which do not intersect with those of y may wellintersect with those of z. As it happens we can do something rathersimple about this, but let's not go there just now. Let's just say thatthe neighbourhood of x that is disjoint from that of y may not bedisjoint from that of z. But if X is Hausdorff there will be aneighbourhood of x, maybe not A, that is disjoint from that of z.

Right. Continuous maps. Recall that f: X → Y is continuous iff for eachx in X, if there is some open set δ in Y containing f(x) there is anopen set ε in X containing x.

Suppose that f is a bijective function. Now if it is the case thatthere is also a continuous map f-¹: Y → X (same definition of continuity), then f is said to be a homeomorphism. I bolded the e there to emphasize this is not a homomorphism. In fact, a little thought suggests it is the topological space equivalent to the concept of isomorphism we meet in group theory.

This is the central theme of topology that I want to get across. Itimplies that, for example, viewed as topological spaces, the circle,the triangle and the rectangle are homeomorphic: one can becontinuously deformed into another and back again, without cutting orgluing. (It gives rise to the joke I'm sure you've all seen about thetopologist who doesn't know the difference between her coffee mug and her doughnut)

Re: Lie groups anyone?

A topological space is said to be connected iff it is not theunion of non-empty disjoint sets. Effecitively this means we can roamover the space without ever falling into a hole.

I'm still trying to get this definition of connected through my head. You say "iff it is not...". What is the "it" in that sentence? S? T? A topological space is a set combined with a set of its subsets, but you talk about it here as if its one set.

Regarding "open" this is slightly more subtle: a subset U of S is openiff it is included in a topology T on S. Note that subsets can be bothopen and closed or neither under this definition.

This definition does seem to comply with any idea of open that I have. Nor do I see how a set could be open, closed, or neither under this definition. Is the definition of closed the negation of this?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Re: Lie groups anyone?

Ricky wrote:

I'm still trying to get this definition of connected through my head. You say "iff it is not...". What is the "it" in that sentence? S? T? A topological space is a set combined with a set of its subsets, but you talk about it here as if its one set.

I'm sorry if I gave the impression we are here dealing with twoseparable entities. We are not. "Combined" is a misleaing way ofputting it.

Look. As you yourself point out in your excellent intro to sets, a set issimply a bunch of things with no intrinsic structure. The notation (S,T) is very rarely used in practice, and merely says we have here aset S with some additional structure denoted by T.

By analogy, recall we agreed a group is a set with an operation. The formal way of denoting this is (G, ·). A vector space is a set V together with itsunderlying field F. We would write (V, F). And so on. These guys arenot the "combination" of two entities. So the "it" is the topologicalspace, a single enity; S specifies what set we are dealing with, Ttells us what the topology on that set is.

But it is a piece of formalism you don't need to worry too much about, we normally say S is a topological space. If we care what particular topology we are dealing with, we would specify it, but usually we don't care.

This definition does seem to comply with any idea of open that I have. Nor do I see how a set could be open, closed, or neither under this definition. Is the definition of closed the negation ofthis?

I take it you meant it does not accord with your idea of open?Well, as I tried to point out, in the topological space R (the realline) with the standard topology i.e. the way we normally think aboutR, open intervals (a, b) are open sets, closed intervals[a, b] are closed sets. The standard topology on R is the union of allopen intervals e.g {(a, c) U (b, d).....}. These intervals are theelements in T, aka open subsets of R. This much you will recognise. So. What about [a, b)? Open, closed, neither, both?

But not all spaces admit of the notion of an interval (technically,this requires a metric; b is in the interval (a, c) iff a < b <c) so we need an alternative definition.

Take for example what I called the discrete topololgy on S = {a, b}, T = {{a}, {b}, {a, b}, Ø}. {a} is in T, and therefore open, {b} is in T, also open. But {b} is the complement of {a} in {a, b} (i.e. {b} = {a, b} - {a}) and is also closed, likewise {a} is the complement of {b}, so it is also closed. So in this topology all subsets are both open and closed.

Note carefully: all topologies on S have as elements S and Ø, so these are open subsets of S. But S is the complement of Ø, and vice versa, so S and Ø are always both open and closed.

I hope you can see this definition of openness includes but is notrestricted to the one we all grew up with. I could say somethingabout set boundaries if you want, but that will require something of adetour. We'll see.

Re: Lie groups anyone?

A topological space is said to be connected iff it is not theunion of non-empty disjoint sets. Effecitively this means we can roamover the space without ever falling into a hole.

I do not believe this definition is correct. (1, 2) and [2, 4) are disjoint and non-empty. But (1, 2) U [2, 4) = (1, 4). Are you telling me that (1, 4) is disconnected?

The standard topology on R is the union of all open intervals e.g {(a, c) U (b, d).....}

Then you are using the "real number" definition of open to define the topology of R. So in other words, the topological definition of open is not an equivalent of the real number definition of open. Correct?

With this, it makes sense if we define the topology of R to be the collection of all open sets.

So. What about [a, b)? Open, closed, neither, both?

Neither, since [a, b) is not in the topology and neither is the complement of [a, b).

Edit: Also, to avoid confusion, please don't go on with new things, as I have some things on post #4, but I want to wait till the above is cleared up.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Re: Lie groups anyone?

Oh? This is a contradiction. Union of disjoint intervals can never generate another interval.

Are you telling me that (1, 4) is disconnected?

I supppose I should blame myself, but the notion of connectedness refers to the whole space, not to particular elements of it.

Then you are using the "real number" definition of open to define the topology of R. So in other words, the topological definition of open is not an equivalent of the real number definition of open. Correct?

Other way round. I am generalizing in order to capture the real line definition as a special case of open so that it applies equally to topological spaces with no sense of order. And yes, the two defintions are equivalent.

With this, it makes sense if we define the topology of R to be the collection of all open sets.

Well union rather than collection, but yes. Good.

So. What about [a, b)? Open, closed, neither, both?

Neither, since [a, b) is not in the topology and neither is the complement of [a, b).

Excellent

Edit: Also, to avoid confusion, please don't go on with new things, as I have some things on post #4, but I want to wait till the above is cleared up.

Re: Lie groups anyone?

Oh? This is a contradiction. Union of disjoint intervals can never generate another interval.

Can you name a real number that is in (1, 4) that is not in (1, 2) U [2, 4)? I believe you are confusing disjoint with separated.

Other way round. I am generalizing in order to capture the real line definition as a special case of open so that it applies equally to topological spaces with no sense of order. And yes, the two defintions are equivalent.

But look at how you defined the topology on R:

The standard topology on R is the union of all open intervals e.g {(a, c) U (b, d).....}

So the topology of R is the union of open intervals, and an interval is open if it is within the topology of R. Seems rather circular.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Re: Lie groups anyone?

Can you name a real number that is in (1, 4) that is not in (1, 2) U [2, 4)? I believe you are confusing disjoint with separated.

The problem here is that there is no topology on R that has both these intervals as open sets (or closed for that matter). Disjoint means that (1, 2) ∩ [2, 4) = Ø, which is clearly the case here.

However, if you don't like my "negative" definition of connected (I promise you it is correct), try this (which I was trying to spare you, as it is less intuitive). A topological space is connected iff the only sets both open and closed are Ø and S (the underlying set).

So the topology of R is the union of open intervals, and an interval is open if it is within the topology of R. Seems rather circular.

No. A set is open if is is an element in the standard topology on R. It so happens that open sets in this topology are open intervals.

The point here is that we start with our intuition about openness in the real line, and try to find a formulation which is true here and also where the notion of an interval makes no sense

Re: Lie groups anyone?

The problem here is that there is no topology on R that has both these intervals as open sets (or closed for that matter). Disjoint means that (1, 2) ∩ [2, 4) = Ø, which is clearly the case here.

However, if you don't like my "negative" definition of connected (I promise you it is correct), try this (which I was trying to spare you, as it is less intuitive). A topological space is connected iff the only sets both open and closed are Ø and S (the underlying set).

Every place I look (including my real analysis book) says that connected is when one set intersected with the closure of the other set is non-empty. In that way:

Another definition from mathworld says that a set is connected if it is not the union of two disjoint open sets. Again, this would mean that (1, 2) U [2, 4) is connected as there aren't two open sets which don't overlap and cover all numbers, so to say.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Re: Lie groups anyone?

Sure, but singletons are always closed (they may in addition be open, but they are never open and not closed) so, first we cannot be sure they are in some topology (we are seeking a general definition here, remember). Second, if they are, as a topological space is connected iff the sets that are both open and closed are S and Ø, then any space with singletons in the topology must be disconnected.

(By the way, if you're using Windows on a PC, use alt+239 (number pad) for ∩ and alt+0472 for Ø. I can send the full list of non-LaTex codes for Windows if anyone wants)

Another definition from mathworld says that a set is connected if it is not the union of two disjoint open sets. Again, this would mean that (1, 2) U [2, 4) is connected as there aren't two open sets which don't overlap and cover all numbers, so to say.

This is perfectly correct, as we expect from Wolfram. But we are not doing set theory! You are again resisting the defintion of openness in the theory of topological spaces. A set is open if it is an element in the topology. Period.

Of course R (standard topology) is connected, that's where this line of thinking starts. But sets of the form (a, b) are in this topology, and sets of the form [c, d) are not, as you said yourself 2 days ago.

Or. In set theory we would say [c, d) is open on the left and closed on the right i.e. [c, d) is both open and closed, right? As it happens there is a topology on R based on sets of this form (look up lower limit topology, or Sorgenfrey line), but we are talking for the moment about the standard topology on R.

Re: Lie groups anyone?

Yes.

OK, I rechecked all my sources, and they all agree that, in the first definition of connectedness I gave, namely that a space is conncted iff it is not the union of disjoint non-empty sets, I should have inserted the qualifier "open sets". Sorry if that caused confusion.

Re: Lie groups anyone?

An interval is open if is doesn't contain its endpoints, as you very well know. If you are leading me make a similar assertion for open sets, forget it. Let me do this instead.

That definition just seems to simple to me. But I guess it works. So open intervals make up the topology of R, and the union of open intervals are also open, which means it agrees with everything I know about R so far.

I have no objections/questions to your last post, but let me go through post 4 in a bit.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Re: Lie groups anyone?

Let X be a topological space, and let x, y, z be points in X. Now X isHausdorff if I can find some subset A containing x and some subset Bcontaining y such that A ∩ B = Ø. Now let z be in the neighbourhood(aka open set) C.

Let z be in the neighbourhood of what? Or is C the neighbourhood fo z?

It need not follow that A ∩ C = Ø, nor that B ∩ C = Ø. In other words,neighbourhoods of x which do not intersect with those of y may wellintersect with those of z. As it happens we can do something rathersimple about this, but let's not go there just now. Let's just say thatthe neighbourhood of x that is disjoint from that of y may not bedisjoint from that of z. But if X is Hausdorff there will be aneighbourhood of x, maybe not A, that is disjoint from that of z.

Assuming Hausdorff, if A ∩ B = Ø, must there exist some neighborhood C such that A ∩ C = Ø and B ∩ C = Ø?

This is the central theme of topology that I want to get across. Itimplies that, for example, viewed as topological spaces, the circle,the triangle and the rectangle are homeomorphic: one can becontinuously deformed into another and back again, without cutting orgluing.

Given a unit sphere, triangle with sides of 1, and square with sides of 1, you're claiming it's possible to set up a homeomorphism between each, right? Can you actually do so? What would the function be?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Given a unit sphere, triangle with sides of 1, and square with sides of 1, you're claiming it's possible to set up a homeomorphism between each, right? Can you actually do so? What would the function be?

Any continuous function f with a continuous inverse. This deserves more thorough treatment than I shall have time for today.

I have no objections/questions to your last post, but let me go through post 4 in a bit.

Of course, take your time. It's tough until you get the feel for it, after that it flows along nicely

Re: Lie groups anyone?

Hmm. I begining to wonder if I'm the right person to be explaining this stuff, as I seem to have obscured a couple of rather simple and very familiar concepts.

Let me try this. Let R be the real line in its familiar form, and let a ≠ b be any two points. We know that the line segment conncting a and b can be infinitely subdivided. This is the Hausdorff property of R.

Now move to R². Here we have to consider the infinite subdivision of lines connecting a, b and c taken pairwise. If this can be done, R² is Hausdorff. If we move to Rⁿ (n > 2) we can continue in this fashion, remembering that we can only make pairwise connections. The formulation for the Hausdorff property I gave captures this rather precisely, with the advantage that it requires no notion of distance between points. C'est tout!

The other thing we like about R is that the line connecting a and b is unbroken. This is the connected propery. Similar considerations to that given for Hausdorff (i.e. omitting all reference to distance, intervals etc) lead to the (corrected) definition I gave.